initial commit / add all books

17 files changed, 834 insertions, 0 deletions

diff --git a/3683/CH19/EX19.1/Ex19_1.sce b/3683/CH19/EX19.1/Ex19_1.sce
new file mode 100644
index 000000000..fcad6fe34
--- /dev/null
+++ b/3683/CH19/EX19.1/Ex19_1.sce
@@ -0,0 +1,41 @@
+fck=15//in MPa

+fy=250//in MPa

+l=4//span, in m

+MF=1.6

+a=MF*20

+D=l*10^3/a//in mm

+W1=(D/10^3)*25//self-weight, in kN/m

+W2=1//floor finish, in kN/m

+W3=2//live load, in kN/m

+W=W1+W2+W3//in kN/m

+Wu=1.5*W//in kN/m

+lef=4.125//in m

+Mu=Wu*lef^2/8//in kN-m

+d=sqrt(Mu*10^6/0.149/fck/10^3)//in mm

+dia=12//assume 12 mm dia bars

+D=d+dia/2+15//<125 mm (assumed value), hence OK

+D=125//in mm

+d=D-dia/2-15//in mm

+//steel

+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast

+a=0.87*fy/0.36/fck/10^3

+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation

+p=0.87*fy*0.416*a

+q=-0.87*fy*d

+r=Mu*10^6

+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm

+s1=1000*0.785*dia^2/Ast//in mm

+s1=105//in mm

+pt=1000*0.785*dia^2/s1/10^3/d*100//in %

+Ads=0.15/100*10^3*D//in sq mm

+//provide 8 mm dia bars

+s2=1000*0.785*8^2/Ads//in mm

+s2=265//in mm

+Vu=Wu*lef/2//in kN

+Tv=Vu*10^3/10^3/d//in MPa

+//for M15 and pt=1

+Tc=0.6//in MPa

+//for solid slabs

+Tc=1.3*Tc//in MPa

+//as Tc>Tv, no shear reinforcement required

+mprintf("Summary of design:\nSlab thickness= %d mm\nCover = 15 mm\nMain steel = 12 mm dia @ %d mm c/c\nDistribution steel = 8 mm dia @ %d mm c/c",D,s1,s2)

diff --git a/3683/CH19/EX19.10/Ex19_10.sce b/3683/CH19/EX19.10/Ex19_10.sce
new file mode 100644
index 000000000..92ebaeca3
--- /dev/null
+++ b/3683/CH19/EX19.10/Ex19_10.sce
@@ -0,0 +1,52 @@
+b=0.2//column width, in m

+D=0.3//column depth, in m

+fck=15//in MPa

+fy=415//in MPa

+P1=600//load on column, in kN

+P2=0.05*P1//weight of footing, in kN

+P=P1+P2//in kN

+Pu=1.5*P//in kN

+q=150//bearing capacity of soil, in kN/sq m

+qu=2*q//ultimate bearing capacity of soil, in kN/sq m

+A=Pu/qu//in sq m

+L=sqrt(A)//assuming footing to be square, in m

+L=1.8//round-off, in m

+p=P1*1.5/L^2//soil pressure, in kN/sq m

+p=277.8//round-off, in kN/sq m

+bc=b/D

+ks=0.5+bc//>1

+ks=1

+Tc=0.25*sqrt(fck)*10^3//in kN/sq m

+Tv=Tc

+//let d be the depth of footing in metres

+//case I: consider greater width of shaded portion in Fig. 19.6 of textbook

+d1=L*(L-b)/2*p/(Tc*L+L*p)//in m

+//case II: refer Fig. 19.7 of textbook; we get a quadratic equation of the form e d^2 + f d + g = 0

+e=p+4*Tc

+f=b*p+D*p+2*(b+D)*Tc

+g=-(L^2-b*D)*p

+d2=(-f+sqrt(f^2-4*e*g))/2/e//in m

+d2=0.35//round-off, in m

+//bending moment consideration, refer Fig. 19.8 of textbook

+Mx=1*((L-b)/2)^2/2*p//in kN-m

+My=1*((L-D)/2)^2/2*p//in kN-m

+d3=sqrt(Mx*10^6/0.138/fck/10^3)//<350 mm, hence OK

+//steel

+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast

+a=0.87*fy/0.36/fck/10^3

+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation

+p=0.87*fy*0.416*a

+q=-0.87*fy*d2*10^3

+r=Mx*10^6

+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm

+Ast=L*Ast//steel required for full width of 1.8 m

+//provide 12 mm dia bars

+dia=12//in mm

+n=Ast/0.785/dia^2//no. of 12 mm dia bars

+n=12//round-off

+Tbd=1.6//in MPa

+Ld=dia*0.87*fy/4/Tbd//in mm

+Ld=677//assume, in mm

+//this length is available from the face of the column in both directions

+D=d2*10^3+dia/2+100//in mm

+mprintf("Summary of design:\nOverall depth of footing=%d mm\nCover=100 mm\nSteel-%d bars of 12 mm dia both ways",D,n)

diff --git a/3683/CH19/EX19.11/Ex19_11.sce b/3683/CH19/EX19.11/Ex19_11.sce
new file mode 100644
index 000000000..72a08499b
--- /dev/null
+++ b/3683/CH19/EX19.11/Ex19_11.sce
@@ -0,0 +1,74 @@
+fck=15//in MPa

+fy=415//in MPa

+phi=30//angle of repose, in degrees

+H=5//height of wall, in m

+B=0.6*H//assume, in m

+T=B/4//assume toe to base ratio as 1:4, in m

+W=16//density of retained earth, in kN/cu m

+Wu=1.5*W//factored load, in kN/cu m

+P=Wu*H^2/2*(1-sind(phi))/(1+sind(phi))//in kN

+M1=P*H/3//in kN-m

+M1=167//round-off, in kN-m

+//bending moment at 2.5 m below the top

+h=2.5//in m

+M2=Wu*h^2/2*(1-sind(phi))/(1+sind(phi))*h/3//in kN-m

+M2=21//round-off, in kN-m

+//thickness of stem (at the base)

+d=sqrt(M1*10^6/0.138/fck/1000)//in mm

+d=285//round-off, in mm

+dia=20//assume 20 mm dia bars

+D1=d+dia/2+25//in mm

+D2=200//thickness at top, in mm

+D3=D2+(D1-D2)*h/H//thickness at 2.5 m below top, in mm 

+d3=sqrt(M2*10^6/0.138/fck/1000)//in mm

+D3=d3+dia/2+25//< 260 mm (provided), hence OK

+D3=260//in mm

+d3=D3-dia/2-25//in mm

+//main steel

+//(a) 5 m below the top

+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast

+a=0.87*fy/0.36/fck/10^3

+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation

+p=0.87*fy*0.416*a

+q=-0.87*fy*d

+r=M1*10^6

+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm

+pt=Ast/1000/d*100//in %

+//provide 20 mm dia bars

+s1=1000*0.785*20^2/Ast//in mm

+s1=155//round-off, in mm

+//(b) 2.5 m below the top

+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast

+a=0.87*fy/0.36/fck/10^3

+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation

+p=0.87*fy*0.416*a

+q=-0.87*fy*d3

+r=M2*10^6

+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm

+Astmin=0.12/100*10^3*D3//in sq mm

+Ast=max(Ast,Astmin)//in sq mm

+//provide 12 mm dia bars

+s2=1000*0.785*12^2/Ast//in mm

+s2=360//round-off, in mm

+//distribution steel

+Ads=0.12/100*10^3*D3//in sq mm

+//provide 8 mm dia bars

+s3=1000*0.785*8^2/Ads//in mm

+s3=160//round-off, in mm

+//check for shear

+Vu=P//in kN

+Tv=Vu*10^3/10^3/d//in MPa

+//for M15 grade concrete and pt=0.71

+Tc=0.54//in MPa

+//as Tc > Tv, no shear reinforcement required

+//development length

+//(a) At the base of stem

+dia=20//in mm

+Tbd=1.6//in MPa

+Ld=dia*0.87*fy/4/Tbd//in mm

+Ld=1130//round-off, in mm

+//(b) At 2.5 m below the top

+dia=12//in mm

+Ld=dia*0.87*fy/4/Tbd//in mm

+Ld=680//round-off, in mm

+mprintf("Summary of design:\nThickness of stem (at base) = %d mm\nThickness of stem at top = %d mm\nRefer Fig. 19.10 of textbook for reinforcement details",D1,D2)

diff --git a/3683/CH19/EX19.12/Ex19_12.sce b/3683/CH19/EX19.12/Ex19_12.sce
new file mode 100644
index 000000000..9a201893b
--- /dev/null
+++ b/3683/CH19/EX19.12/Ex19_12.sce
@@ -0,0 +1,15 @@
+P=1000//in kN

+Pu=1.5*P//in kN

+fck=15//in MPa

+fy=415//in MPa

+l=3.5//unsupported length, in m

+//assume 1% steel

+Ag=Pu*10^3/(0.4*fck*0.99+0.67*fy*0.01)//in sq mm

+L=sqrt(Ag)//assuming a square column

+L=420//in mm

+emin=l*10^3/500+L/30//in mm

+ep=0.05*L//=emin, hence OK

+Asc=0.01*L^2//in sq mm

+//provide 6-20 mm dia bars

+Asc=6*0.785*20^2//in sq mm

+mprintf("Summary of design:\nColumn size - %d x %d mm\nSteel-main = 6-20 mm dia bars",L,L)

diff --git a/3683/CH19/EX19.13/Ex19_13.sce b/3683/CH19/EX19.13/Ex19_13.sce
new file mode 100644
index 000000000..6c782e0bb
--- /dev/null
+++ b/3683/CH19/EX19.13/Ex19_13.sce
@@ -0,0 +1,23 @@
+P=500//in kN

+Pu=1.5*P//in kN

+fck=15//in MPa

+fy=250//in MPa

+l=3//unsupported length, in m

+//assume 1% steel

+Ag=Pu*10^3/(0.4*fck*0.99+0.67*fy*0.01)//in sq mm

+L=sqrt(Ag)//assuming a square column

+L=315//in mm

+emin=l*10^3/500+L/30//<20

+emin=20//in mm

+ep=0.05*L//<emin, hence the column is to be checked for bending

+Mu=Pu*10^3*emin//in N-mm

+a=Pu*10^3/fck/L/L

+b=Mu/fck/L/L^2//b=0.032

+d1=40//cover(assume), in mm

+c=d1/L//c=d'/D

+//for d'/D = 0.15

+p=0.07*fck//in %

+Asc=p/100*L^2//in sq mm

+//provide 4-20 mm dia bars

+Asc=4*0.785*20^2//in sq mm

+mprintf("Summary of design:\nColumn size - %d x %d mm\nSteel-main = 4-20 mm dia bars",L,L)

diff --git a/3683/CH19/EX19.14/Ex19_14.sce b/3683/CH19/EX19.14/Ex19_14.sce
new file mode 100644
index 000000000..063d07d44
--- /dev/null
+++ b/3683/CH19/EX19.14/Ex19_14.sce
@@ -0,0 +1,30 @@
+P=500//in kN

+Pu=1.5*P//in kN

+fck=15//in MPa

+fy=250//in MPa

+l=3//unsupported length, in m

+//assume 1% steel

+Ag=Pu*10^3/(0.4*fck*0.99+0.67*fy*0.01)//in sq mm

+b=250//in mm

+D=Ag/b//in mm

+D=400//round-off, in mm

+emin1=l*10^3/500+D/30//in direction of Y axis, in mm, < 20 mm

+emin1=20//in mm

+ep1=0.05*D//=emin, hence no moment is required to be considered in this direction

+emin2=l*10^3/500+b/30//in direction of X axis, in mm, < 20 mm

+emin2=20//in mm

+ep2=0.05*b//<emin, hence moment in this direction needs to be considered

+//interaction diagram

+b=400//in mm

+D=250//in mm

+Mu=Pu*10^3*emin2//in N-mm

+m=Pu*10^3/fck/b/D

+n=Mu/fck/b/D^2//b=0.032

+d1=40//cover(assume), in mm

+c=d1/D//c=d'/D

+//referring to Fig. 19.12

+p=0.08*fck//in %

+Asc=p/100*b*D//in sq mm

+//provide 6-16 dia bars

+Asc=6*0.785*16^2//in sq mm

+mprintf("Summary of design:\nColumn size - %d x %d mm\nSteel-main = 6-16 mm dia bars",D,b)

diff --git a/3683/CH19/EX19.15/Ex19_15.sce b/3683/CH19/EX19.15/Ex19_15.sce
new file mode 100644
index 000000000..b57d28ca9
--- /dev/null
+++ b/3683/CH19/EX19.15/Ex19_15.sce
@@ -0,0 +1,32 @@
+P=500//in kN

+Pu=1.5*P//in kN

+fck=15//in MPa

+fy=250//in MPa

+l=5//effective length, in m

+lex=5//in m

+ley=5//in m

+L=315//column dimension in mm (square column)

+Asc=1256//in sq mm

+m=lex*10^3/L//>12

+n=ley*10^3/L//>12

+//hence the column is slender on both the axes

+Max=Pu*10^3*L/2000*(lex/(L/10^3))^2/10^6//in kN-m

+May=Max

+Puz=(0.45*fck*(L^2-Asc)+0.75*fy*Asc)/10^3//in kN

+c=40//cover, in mm

+//to find Pb

+xu=(L-c)/(1+0.002/0.0035)//in mm

+Pb=0.36*fck*L*xu/10^3//in kN

+k=(Puz-Pu)/(Puz-Pb)//>1

+Max=k*Max//in kN-m

+Mu=15//in kN-m

+Mu=Mu+Max//in kN-m

+a=Pu*10^3/fck/L/L

+b=Mu*10^6/fck/L/L^2//b=0.047

+d1=c/L//d1=d'/D

+//for d'/D = 0.1

+p=0.095*fck//in %

+Asc=p/100*L^2//in sq mm

+//provide 4-18 mm + 4-12 mm dia bars

+Asc=4*0.785*18^2+4*0.785*12^2//in sq mm

+mprintf("Summary of design:\nColumn size - %d x %d mm\nSteel-main = 4-18 mm + 4-12 mm dia bars",L,L)

diff --git a/3683/CH19/EX19.16/Ex19_16.sce b/3683/CH19/EX19.16/Ex19_16.sce
new file mode 100644
index 000000000..8e3a75cf6
--- /dev/null
+++ b/3683/CH19/EX19.16/Ex19_16.sce
@@ -0,0 +1,37 @@
+Pu=2000//in kN

+Mux=50//in kN-m

+Muy=Mux

+fck=20//in MPa

+fy=415//in MPa

+//assume 2% steel

+p=2//in %

+Ag=Pu*10^3/(0.4*fck*(1-p/100)+0.67*fy*p/100)//in sq mm

+L=sqrt(Ag)//assuming a square column

+L=400//in mm

+m=Pu*10^3/fck/L/L

+n=p/fck

+c=50//cover (assume), in mm

+d1=c/L//d1=d'/D

+//from Fig. 19.21, for d'/D = 0.15 and Pu / fck b D = 0.625

+f=0.046

+Mux1=f*fck*L*L^2/10^6//in kN-m

+Muy1=Mux1

+Puz=(0.45*fck*(1-p/100)*L^2+0.75*fy*p/100*L^2)/10^3//in kN

+a=Pu/Puz//>0.8

+an=2

+b=(Mux/Mux1)^an+(Muy/Muy1)^an//>1

+//assume 2.5% steel

+p=2.5//in %

+n=p/fck

+//from Fig. 19.21, for d'/D = 0.15 and Pu / fck b D = 0.625

+f=0.08

+Mux1=f*fck*L*L^2/10^6//in kN-m

+Muy1=Mux1

+Puz=(0.45*fck*(1-p/100)*L^2+0.75*fy*p/100*L^2)/10^3//in kN

+a=Pu/Puz//<0.8

+an=1+1/0.6*(a-0.2)

+b=(Mux/Mux1)^an+(Muy/Muy1)^an//<1, hence OK

+Asc=p/100*L^2//in sq mm

+//provide 12-22 mm dia bars

+Asc=12*0.785*22^2//in sq mm

+mprintf("Summary of design:\nColumn size - %d x %d mm\nSteel-main = 12-22 mm dia bars placed equally on four faces of the column",L,L)

diff --git a/3683/CH19/EX19.17/Ex19_17.sce b/3683/CH19/EX19.17/Ex19_17.sce
new file mode 100644
index 000000000..c6a8814da
--- /dev/null
+++ b/3683/CH19/EX19.17/Ex19_17.sce
@@ -0,0 +1,33 @@
+b=400//in mm

+D=500//in mm

+Pu=1600//in kN

+Mux=90//in kN-m

+Muy=50//in kN-m

+fck=15//in MPa

+fy=415//in MPa

+p=1.5//assume 1.5% steel, placed on four sides

+m=p/fck

+c=50//cover (assume), in mm

+//to find Mux1

+n=c/D//n=d'/D

+l=Pu*10^3/fck/b/D

+//referring to Fig.19.20, for Pu/ fck/ b/ D = 0.53 and p/ fck = 0.1

+f=0.09

+Mux1=f*fck*b*D^2/10^6//in kN-m

+//to find Muy1

+b=500//in mm

+D=400//in mm

+n=c/D//n=d'/D

+l=Pu*10^3/fck/b/D

+//referring to Fig.19.21, for Pu/ fck/ b/ D = 0.53 and p/ fck = 0.1

+f=0.08

+Muy1=f*fck*b*D^2/10^6//in kN-m

+Puz=(0.45*fck*(1-p/100)*b*D+0.75*fy*p/100*b*D)/10^3//in kN

+a=Pu/Puz//<0.8

+an=1+1/0.6*(a-0.2)

+r=(Mux/Mux1)^an+(Muy/Muy1)^an//<1

+Asc=p/100*b*D//in sq mm

+//provide 6-16 mm + 6-20 mm dia bars

+Asc=6*0.785*16^2+6*0.785*20^2//in sq mm

+mprintf("Summary of design:\nColumn size - %d x %d mm\nSteel-main = 6-16 mm + 6-20 mm dia bars",D,b)

+//answer in textbook is incorrect

diff --git a/3683/CH19/EX19.18/Ex19_18.sce b/3683/CH19/EX19.18/Ex19_18.sce
new file mode 100644
index 000000000..27dd70c0a
--- /dev/null
+++ b/3683/CH19/EX19.18/Ex19_18.sce
@@ -0,0 +1,73 @@
+b=300//in mm

+Pu=1500//in kN

+Mux=100//in kN-m

+Muy=70//in kN-m

+fck=15//in MPa

+fy=250//in MPa

+p=1.5//assume 1.5% steel, placed on four sides

+Ag=Pu*10^3/(0.4*fck*(1-p/100)+0.67*fy*p/100)//in sq mm

+D=Ag/b//in mm

+D=600//assume, in mm

+m=p/fck

+c=60//cover (assume), in mm

+//to find Mux1

+n=c/D//n=d'/D

+l=Pu*10^3/fck/b/D

+//referring to Fig.19.17, for Pu/ fck/ b/ D = 0.56 and p/ fck = 0.1

+f=0.038

+Mux1=f*fck*b*D^2/10^6//in kN-m

+//to find Muy1

+b=600//in mm

+D=300//in mm

+n=c/D//n=d'/D

+l=Pu*10^3/fck/b/D

+//referring to Fig.19.19, for Pu/ fck/ b/ D = 0.56 and p/ fck = 0.1

+f=0.038

+Muy1=f*fck*b*D^2/10^6//in kN-m

+Puz=(0.45*fck*(1-p/100)*b*D+0.75*fy*p/100*b*D)/10^3//in kN

+a=Pu/Puz//>0.8

+an=2

+r=(Mux/Mux1)^an+(Muy/Muy1)^an//>1

+p=4//assume 4% steel, second trial

+m=p/fck

+//to find Mux1

+b=300//in mm

+D=600//in mm

+//referring to Fig.19.17, for Pu/ fck/ b/ D = 0.56 and p/ fck = 0.26

+f=0.15

+Mux1=f*fck*b*D^2/10^6//in kN-m

+//to find Muy1

+b=600//in mm

+D=300//in mm

+n=c/D//n=d'/D

+//referring to Fig.19.19, for Pu/ fck/ b/ D = 0.56 and p/ fck = 0.26

+f=0.15

+Muy1=f*fck*b*D^2/10^6//in kN-m

+Puz=(0.45*fck*(1-p/100)*b*D+0.75*fy*p/100*b*D)/10^3//in kN

+a=Pu/Puz//<0.8

+an=1+1/0.6*(a-0.2)

+r=(Mux/Mux1)^an+(Muy/Muy1)^an//<1, hence OK

+//but steel can be reduced

+p=3//assume 3% steel, second trial

+m=p/fck

+//to find Mux1

+b=300//in mm

+D=600//in mm

+//referring to Fig.19.17, for Pu/ fck/ b/ D = 0.56 and p/ fck = 0.2

+f=0.12

+Mux1=f*fck*b*D^2/10^6//in kN-m

+//to find Muy1

+b=600//in mm

+D=300//in mm

+n=c/D//n=d'/D

+//referring to Fig.19.19, for Pu/ fck/ b/ D = 0.56 and p/ fck = 0.2

+f=0.12

+Muy1=f*fck*b*D^2/10^6//in kN-m

+Puz=(0.45*fck*(1-p/100)*b*D+0.75*fy*p/100*b*D)/10^3//in kN

+a=Pu/Puz//<0.8

+an=1+1/0.6*(a-0.2)

+r=(Mux/Mux1)^an+(Muy/Muy1)^an//<1, hence OK

+Asc=p/100*b*D//in sq mm

+//provide 12-25 dia bars

+Asc=12*0.785*25^2//in sq mm

+mprintf("Summary of design:\nColumn size - %d x %d mm\nSteel-main = 12-25 mm dia bars",D,b)

diff --git a/3683/CH19/EX19.3/Ex19_3.sce b/3683/CH19/EX19.3/Ex19_3.sce
new file mode 100644
index 000000000..33876dac8
--- /dev/null
+++ b/3683/CH19/EX19.3/Ex19_3.sce
@@ -0,0 +1,43 @@
+fck=15//in MPa

+fy=415//in MPa

+l=4.5//span, in m

+MF=1.4

+a=MF*20

+D=l*10^3/a//in mm

+D=160//in mm

+W1=(D/10^3)*25//self-weight, in kN/m

+W2=1//floor finish, in kN/m

+W3=1//partitions, in kN/m

+W4=4//live load, in kN/m

+W=W1+W2+W3+W4//in kN/m

+Wu=1.5*W//in kN/m

+lef=l+0.16//in m

+Mu=Wu*lef^2/8//in kN-m

+d=sqrt(Mu*10^6/0.138/fck/10^3)//in mm

+dia=12//assume 12 mm dia bars

+D=d+dia/2+15//=160 mm(assumed value), approximately

+D=160//in mm

+d=140//in mm

+//steel

+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast

+a=0.87*fy/0.36/fck/10^3

+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation

+p=0.87*fy*0.416*a

+q=-0.87*fy*d

+r=Mu*10^6

+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm

+s1=1000*0.785*dia^2/Ast//in mm

+s1=112//in mm

+pt=Ast/10^3/d*100//in %

+Ads=0.12/100*10^3*D//in sq mm

+//provide 8 mm dia bars

+s2=1000*0.785*8^2/Ads//in mm

+s2=260//in mm

+Vu=Wu*lef/2//in kN

+Tv=Vu*10^3/10^3/d//in MPa

+//for M15 and pt=0.718

+Tc=0.53//in MPa

+//for solid slabs

+Tc=1.25*Tc//in MPa

+//as Tc>Tv, no shear reinforcement required

+mprintf("Summary of design:\nSlab thickness= %d mm\nCover = 15 mm\nMain steel = 12 mm dia @ %d mm c/c\nDistribution steel = 8 mm dia @ %d mm c/c",D,s1,s2)

diff --git a/3683/CH19/EX19.4/Ex19_4.sce b/3683/CH19/EX19.4/Ex19_4.sce
new file mode 100644
index 000000000..430eca029
--- /dev/null
+++ b/3683/CH19/EX19.4/Ex19_4.sce
@@ -0,0 +1,48 @@
+fck=15//in MPa

+fy=415//in MPa

+MF=1.4//modification factor

+//let a be span to depth ratio

+l=1//span, in m

+a=MF*7

+D=l*1000/a//in mm

+D=105//assume, in mm

+//to calculate loading

+W1=25*(D/10^3)*1.5//self-weight, in kN/m

+W2=0.5*1.5//finish, in kN/m

+W3=0.75*1.5//live load, in kN/m

+W=W1+W2+W3//in kN/m

+Wu=1.5*W//in kN/m

+lef=l+0.23/2//effective span, in m

+Mu=Wu*lef/2//in kN-m

+//check for depth

+d=sqrt(Mu*10^6/(0.138*fck*1500))//in mm

+dia=12//assume 12 mm dia bars

+D=d+12/2+15//<105, hence OK

+D=100//assume, in mm

+d=D-dia/2-15//in mm

+//steel

+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast

+a=0.87*fy/0.36/fck/1.5/10^3

+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation

+p=0.87*fy*0.416*a

+q=-0.87*fy*d

+r=Mu*10^6

+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm

+//provide 8 mm dia bars

+dia=8//in mm

+s1=1500*0.785*dia^2/Ast//>3d=3x79=237 mm

+s1=235//in mm

+Ads=0.12/100*1000*D//distribution steel, in sq mm

+//assume 6 mm dia bars

+s2=1000*0.785*6^2/Ads//in mm

+s2=235//round-off, in mm

+Tbd=1.6//in MPa

+Ld=dia*0.87*fy/4/Tbd//in mm

+Ld=452//in mm

+Tv=Wu*10^3/1500/d//in MPa

+Ast=1500*0.785*8^2/235//in sq mm

+pt=Ast/1500/d*100//in %

+//for M15 and pt=0.26

+Tc=0.35//in MPa

+//as Tc>Tv, no shear reinforcement required

+mprintf("Summary of design\nThickness of slab = %d mm\nCover = 15 mm\nMain steel = 8 mm dia @ %d mm c/c\nDevelopment length = %d mm\nDistribution steel = 6 mm dia @ %d mm c/c",D,s1,Ld,s2)

diff --git a/3683/CH19/EX19.5/Ex19_5.sce b/3683/CH19/EX19.5/Ex19_5.sce
new file mode 100644
index 000000000..6c01260cc
--- /dev/null
+++ b/3683/CH19/EX19.5/Ex19_5.sce
@@ -0,0 +1,43 @@
+lx=3.5//in m

+ly=4//in m

+fck=15//in MPa

+fy=250//in MPa

+D=lx*10^3/35//in mm

+W1=(D/10^3)*25//self-weight, in kN/m

+W2=1.5//live load, in kN/m

+W=W1+W2//in kN/m

+Wu=1.5*W//in kN/m

+a=ly/lx

+Ax=0.078

+Ay=0.0602

+Mx=Ax*Wu*lx^2//in kN-m

+My=Ay*Wu*lx^2//in kN-m

+d=sqrt(Mx*10^6/0.149/fck/10^3)//in mm

+d=51//round-off, in mm

+//assume 10 mm dia bars

+dia=10//in mm

+D=d+dia/2+15//<100 mm assumed value

+D=100//in mm

+d=D-dia/2-15//in mm

+//steel - short span

+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast

+a=0.87*fy/0.36/fck/10^3

+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation

+p=0.87*fy*0.416*a

+q=-0.87*fy*d

+r=Mx*10^6

+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm

+s1=1000*0.785*dia^2/Ast//in mm

+s1=220//round-off, in mm

+//long span

+d=d-dia/2-dia/2//in mm

+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast

+a=0.87*fy/0.36/fck/10^3

+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation

+p=0.87*fy*0.416*a

+q=-0.87*fy*d

+r=My*10^6

+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm

+s2=1000*0.785*dia^2/Ast//in mm

+s2=250//round-off, in mm

+mprintf("Summary of design\nSlab thickness=%d mm\nCover=15 mm\nSteel-\n(i)Short span = 10 mm dia @ %d mm c/c\n(ii)Long span = 10 mm dia @ %d mm c/c",D,s1,s2)

diff --git a/3683/CH19/EX19.6/Ex19_6.sce b/3683/CH19/EX19.6/Ex19_6.sce
new file mode 100644
index 000000000..c9f8bea98
--- /dev/null
+++ b/3683/CH19/EX19.6/Ex19_6.sce
@@ -0,0 +1,79 @@
+b=225//width in mm

+D=300//depth in mm

+fck=15//in MPa

+fy=415//in MPa

+l=4.2//span, in m

+W1=(b/10^3)*(D/10^3)*25//self-weight, in kN/m

+W2=6//live load, in kN/m

+W=W1+W2//in kN/m

+Wu=1.5*W//in kN/m

+Mu=Wu*l^2/8//in kN-m

+d=270//assume, in mm

+Mulim=0.138*fck*b*d^2/10^6//in kN-m

+//as Mulim > Mu, it will be a singly reinforced beam

+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast

+a=0.87*fy/0.36/fck/b

+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation

+p=0.87*fy*0.416*a

+q=-0.87*fy*d

+r=Mu*10^6

+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm

+//provide 12 mm dia bars

+n=Ast/0.785/12^2

+n=3//assume

+Ast=n*0.785*12^2//in sq mm

+Vu=Wu*l/2//in kN

+Tv=Vu*10^3/b/d//in MPa

+pt=Ast/b/d*100//pt=0.56

+//for M15 and pt=0.56

+Tc=0.46//in MPa

+//as Tc>Tv, no shear reinforcement required

+//provide nominal stirrups and provide 6 mm stirrups

+Asv=2*0.785*6^2//in sq mm

+Sv=Asv*fy/0.4/b//in mm

+Sv=260//assume, in mm

+Svmax=0.75*d//in mm

+Svmax=200//round-off, in mm

+Sv=min(Sv,Svmax)//in mm

+mprintf("Summary of design:\nBeam size - %d x %d mm\nCover - 25 mm\nSteel - %d-12 mm dia bars\nStirrups - 6 mm dia @ %d mm c/c",b,D,n,Sv)

+//deflection check

+Ec=5700*sqrt(fck)//in MPa

+Es=2*10^5//in MPa

+m=Es/Ec

+fcr=0.7*sqrt(fck)//in MPa

+//using b x x/2 = m Ast (d-x), we get a quadratic equation

+//solving the quadratic equation

+p=b/2

+q=m*Ast

+r=-m*Ast*d

+x=(-q+sqrt(q^2-4*p*r))/2/p//in mm

+z=d-x/3//in mm

+Ir=b*x^3/12+b*x*(x/2)^2+m*Ast*(d-x)^2//in mm^4

+Igr=b*D^3/12//in mm^4

+yt=D/2//in mm

+Mr=fcr*Igr/yt//in N-mm

+M=W*l^2/8*10^6//in N-mm

+Ieff=Ir/(1.2-Mr/M*z/d*(1-x/d)*b/b)//in mm^4

+//Ir<Ieff<Igr, hence OK

+W1=W*l//in kN

+u1=5/384*(W1*10^3)*(l*10^3)^3/Ec/Ieff//short-term deflection, in mm

+//long-term deflection

+//(i) deflection due to shrinkage

+k3=0.125//for simply supported beam

+pt=0.56//in %

+pc=0//in %

+k4=0.72*(pt-pc)/sqrt(pt)

+phi=k4*0.0003/D

+u2=k3*phi*(l*10^3)^2//in mm

+//(ii) deflection due to creep

+Ecc=Ec/(1+1.6)//in MPa

+//assuming a permanent load of 60%

+W2=0.6*W*l//in kN

+u3=5/384*(W2*10^3)*(l*10^3)^3/Ecc/Ieff//in mm

+u4=5/384*(W2*10^3)*(l*10^3)^3/Ec/Ieff//in mm

+u5=u3-u4//in mm

+u=u1+u2+u5//total deflection, in mm

+v1=l*10^3/250//permissible deflection, in mm

+v2=l*10^3/350//in mm

+//assuming half the shrinkage strain occurs within the first 28 days, the deflection occurring after this time

+v3=u2/2+u5//< permissible value, hence OK

diff --git a/3683/CH19/EX19.7/Ex19_7.sce b/3683/CH19/EX19.7/Ex19_7.sce
new file mode 100644
index 000000000..de5c0cd3e
--- /dev/null
+++ b/3683/CH19/EX19.7/Ex19_7.sce
@@ -0,0 +1,89 @@
+l=7//span, in m

+fck=15//in MPa

+fy=250//in MPa

+b=300//assume, in mm

+W=35//live load, in kN/m

+Wu=1.5*W//in kN/m

+Mu=Wu*l^2/8//in kN-m

+d=(Mu*10^6/0.149/fck/b)^0.5//in mm

+d=1.1*d//increase depth by 10% for self-weight

+d=750//assume, in mm

+c=50//cover, in mm

+D=d+c//in mm

+W1=(b/10^3)*(D/10^3)*25//self-weight, in kN/m

+W2=35//live load, in kN/m

+W=W1+W2//in kN/m

+Wu=1.5*W//in kN/m

+Mu=Wu*l^2/8//in kN-m

+d=(Mu*10^6/0.149/fck/b)^0.5//<750 mm, hence OK

+d=750//in mm

+//steel

+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast

+a=0.87*fy/0.36/fck/b

+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation

+p=0.87*fy*0.416*a

+q=-0.87*fy*d

+r=Mu*10^6

+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm

+//provide 20 mm dia bars

+n=Ast/0.785/20^2

+//provide 8-20 mm + 2-18 mm dia bars

+Ast=8*0.785*20^2+2*0.785*18^2//in sq mm

+pt=Ast/b/d*100//pt=1.34

+Vu=Wu*l/2//in kN

+Tv=Vu*10^3/b/d//in MPa

+//for M15 and pt=1.34

+Tc=0.65//in MPa

+//as Tv>Tc, shear reinforcement required

+//provide 6 mm stirrups

+Vus=Vu-Tc*b*d/10^3//in kN

+Asv=2*0.785*6^2//in sq mm

+Sv=Asv*0.87*fy*d/Vus/10^3//in mm

+Sv=130//assume, in mm

+Svmin=Asv*fy/0.4/b//in mm

+Svmin=115//assume, in mm

+Sv=min(Sv,Svmin)//in mm

+mprintf("Summary of design:\nBeam size - %d x %d mm\nCover - 50 mm\nSteel - 8-20 mm + 2-18 mm dia bars\nStirrups - 6 mm dia @ %d mm c/c",b,D,Sv)

+//deflection check

+Ec=5700*sqrt(fck)//in MPa

+Es=2*10^5//in MPa

+m=Es/Ec

+fcr=0.7*sqrt(fck)//in MPa

+//using b x x/2 = m Ast (d-x), we get a quadratic equation

+//solving the quadratic equation

+p=b/2

+q=m*Ast

+r=-m*Ast*d

+x=(-q+sqrt(q^2-4*p*r))/2/p//in mm

+x=290//assume, in mm

+z=d-x/3//in mm

+Ir=b*x^3/12+b*x*(x/2)^2+m*Ast*(d-x)^2//in mm^4

+Igr=b*D^3/12//in mm^4

+yt=D/2//in mm

+Mr=fcr*Igr/yt//in N-mm

+M=W*l^2/8*10^6//in N-mm

+Ieff=Ir/(1.2-Mr/M*z/d*(1-x/d)*b/b)//in mm^4

+//Ir>Ieff

+Ieff=Ir//in mm^4

+W1=W*l//in kN

+u1=5/384*(W1*10^3)*(l*10^3)^3/Ec/Ieff//short-term deflection, in mm

+//long-term deflection

+//(i) deflection due to shrinkage

+k3=0.125//for simply supported beam

+pt=1.34//in %

+pc=0//in %

+k4=0.65*(pt-pc)/sqrt(pt)

+phi=k4*0.0003/D

+u2=k3*phi*(l*10^3)^2//in mm

+//(ii) deflection due to creep

+Ecc=Ec/(1+1.6)//in MPa

+//assuming a permanent load of 60%

+W2=0.6*W*l//in kN

+u3=5/384*(W2*10^3)*(l*10^3)^3/Ecc/Ieff//in mm

+u4=5/384*(W2*10^3)*(l*10^3)^3/Ec/Ieff//in mm

+u5=u3-u4//in mm

+u=u1+u2+u5//total deflection, in mm

+v1=l*10^3/250//permissible deflection, in mm

+v2=l*10^3/350//in mm

+//assuming half the shrinkage strain occurs within the first 28 days, the deflection occurring after this time

+v3=u2/2+u5//< permissible value, hence OK

diff --git a/3683/CH19/EX19.8/Ex19_8.sce b/3683/CH19/EX19.8/Ex19_8.sce
new file mode 100644
index 000000000..c9609ce52
--- /dev/null
+++ b/3683/CH19/EX19.8/Ex19_8.sce
@@ -0,0 +1,86 @@
+l=10//span, in m

+fck=15//in MPa

+fy=250//in MPa

+Df=100//slab thickness, in mm 

+D=l*10^3/15//depth of beam, in mm

+D=600//assume, in mm

+d=D-50//cover=50 mm

+bw=300//beam width, in mm

+bf=l*10^3/6+bw+6*Df//>2500 mm c/c distance of beams

+bf=2500//in mm

+W1=(bw/10^3)*(D-Df)/10^3*25//web, in kN/m

+W2=(Df/10^3)*(bf/10^3)*25//slab, in kN/m

+W3=(bf/10^3)*5//imposed load, in kN/m

+W=W1+W2+W3//in kN/m

+Wu=1.5*W//in kN/m

+Mu=Wu*l^2/8//in kN-m

+Vu=Wu*l/2//in kN

+Asf=0.36*fck*bf*Df/0.87/fy//steel required only for flange, in sq mm

+Asf=6210//round-off, in sq mm

+//verification of trial section

+xu=100//assume, in mm

+Ast=Asf//in sq mm

+Mulim=0.87*fy*Ast*(d-0.416*xu)/10^6//in kN-m

+//Mulim > Mu, hence OK

+//keeping the assumed trial section, work out the steel required

+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast

+a=0.87*fy/0.36/fck/bf

+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation

+p=0.87*fy*0.416*a

+q=-0.87*fy*d

+r=Mu*10^6

+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm

+//provide 5-25 mm dia + 3-22 mm dia bars

+pt=Ast*100/(bw*d+(bf-bw)*Df)//pt=1%, approximately

+//check for shear

+Tv=Vu*10^3/bw/d//in MPa

+//for M15 grade concrete and pt=1%

+Tc=0.6//in MPa

+//as Tv > Tc, shear reinforcement required

+Vus=Vu-Tc*bw*d/10^3//in kN

+//provide 6 mm dia stirrups

+Asv=2*0.785*6^2//in sq mm

+Sv=Asv*0.87*fy*d/Vus/10^3//in mm

+Sv=90//round-off, in mm

+mprintf("T beam:bf=%d mm\nDf=%d mm\nd=%d mm\nD=%d mm\nCover = 50 mm\nSteel= 5-25 mm dia + 3-22 mm dia bars\nStirrups = 6 mm dia @ %d mm c/c throughout",bf,Df,d,D,Sv)

+//answer in textbook for spacing of stirrups is incorrect

+//deflection check

+Ec=5700*sqrt(fck)//in MPa

+Es=2*10^5//in MPa

+m=Es/Ec//modular ratio

+fcr=0.7*sqrt(fck)//in MPa

+//using bf Df (x-Df/2) = m Ast (d-x), we get a quadratic equation

+x=(m*Ast*d+bf*Df^2/2)/(bf*Df+m*Ast)//in mm

+z=0.87*d//assume, in mm

+//refer Fig. 19.5 of textbook

+Ir=bf*x^3/12+bf*Df*(x/2)^2+m*Ast*(d-x)^2//in mm^4

+y=(bf*Df*Df/2+(D-Df)*bw*((D-Df)/2+Df))/(bf*Df+(D-Df)*bw)//c.g. from top, in mm (neglecting steel)

+Igr=bf*Df^3/12+bf*Df*(Df/2-y)^2+bw*(D-Df)^3/12+bw*(D-Df)*((D-Df)/2+Df-y)^2//in mm^4

+yt=d/2//in mm

+Mr=fcr*Igr/yt//in N-mm

+M=W*l^2/8*10^6//in N-mm

+Ieff=Ir/(1.2-Mr/M*z/d*(1-x/d)*bw/bf)//in mm^4

+//Ir > Ieff

+Ieff=Ir//in mm^4

+W1=W*l//in kN

+u1=5/384*(W1*10^3)*(l*10^3)^3/Ec/Ieff//short term deflection, in mm

+//deflection due to shrinkage

+k3=0.125//for simply supported beam

+pt=1//in %

+pc=0//in %

+k4=0.65*(pt-pc)/sqrt(pt)

+phi=k4*0.0003/D

+u2=k3*phi*(l*10^3)^2//in mm

+//deflection due to creep

+Ecc=Ec/(1+1.6)//in MPa

+//assuming a permanent load of 60%

+W2=0.6*W*l//in kN

+u3=5/384*(W2*10^3)*(l*10^3)^3/Ecc/Ieff//in mm

+u4=5/384*(W2*10^3)*(l*10^3)^3/Ec/Ieff//in mm

+u5=u3-u4//in mm

+u=u1+u2+u5//total deflection, in mm

+v1=l*10^3/250//permissible deflection, in mm

+v2=l*10^3/350//>20 mm

+v2=20//in mm

+//assuming half the shrinkage strain occurs within the first 28 days, the deflection occurring after this time

+v3=u2/2+u5//< permissible value, hence OK

diff --git a/3683/CH19/EX19.9/Ex19_9.sce b/3683/CH19/EX19.9/Ex19_9.sce
new file mode 100644
index 000000000..39a5c515b
--- /dev/null
+++ b/3683/CH19/EX19.9/Ex19_9.sce
@@ -0,0 +1,36 @@
+l=2.7+1//span, in m

+R=0.15//rise, in m

+t=0.27//tread, in m

+fck=15//in MPa

+fy=415//in MPa

+D=200//assume, in mm

+W1=D/10^3*25*sqrt(R^2+t^2)/t//slab load on plan, in kN/m

+W2=1/2*R*t*25/t//load of step per metre, in kN/m

+W3=3//live load, in kN/m

+W=W1+W2+W3//in kN/m

+Wu=1.5*W//in kN/m

+Mu=Wu*l^2/8//in kN-m

+d=sqrt(Mu*10^6/0.138/fck/10^3)//in mm

+d=115//round-off, in mm

+//assume 10 mm dia bars

+dia=10//in mm

+D=d+dia/2+25//< 200 mm, hence OK

+D=l*10^3/24//depth required for deflection, in mm

+D=155//round-off, in mm

+d=D-dia/2-25//in mm

+//steel

+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast

+a=0.87*fy/0.36/fck/10^3

+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation

+p=0.87*fy*0.416*a

+q=-0.87*fy*d

+r=Mu*10^6

+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm

+s1=1000*0.785*dia^2/Ast//spacing of 10 mm dia bars

+s1=110//round-off, in mm

+Ads=0.12/100*D*10^3//distribution steel, in sq mm

+//provide 8 mm dia bars

+s2=1000*0.785*8^2/Ads//in mm

+s2=270//round-off, in mm

+mprintf("Summary of design\nSlab thickness=%d mm\nCover = 25 mm\nMain steel = 10 mm dia bars @ %d mm c/c\nDistribution steel = 8 mm dia @ %d mm c/c",D,s1,s2)

+//answer in textbook for spacing of 10 mm dia bars is incorrect